Question: Let $a(x)=8x^3+35x^2-17x-5$, and $b(x)=x^2+5x+3$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Solution: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{8x^3+35x^2-17x-5}{x^2+5x+3}$ : We divide ${x^2}$ into ${8x^3}$ to get ${8x}$ : $ \hphantom{1567|144474} {8x}\\ {{{x^2}+5x+3}}|\overline{{8x^3}+35x^2-17x-5}\\ \hphantom{37...8........|}\llap{-}\underline{(8x^3+ 40x^2 +24x)}\\ \hphantom{37|3....998..........}-5x^2-41x \\ $ [What did we do here?] Next, we divide ${x^2}$ into ${-5x^2}$ to get ${-5}$ : $ \hphantom{1567|166664} {8x \ \ {- \ \ 5}}\\ {{{x^2}+5x+3}}|\overline{8x^3+35x^2-17x-5}\\ \hphantom{37...8........|}\llap{-}\underline{(8x^3+ 40x^2 +24x)}\\ \hphantom{37|3....998..........}{-5x^2}-41x -5\\ \hphantom{37.......888.........|}\llap{-}\underline{(-5x^2-25x-15)}\\ \hphantom{37|3.............777777......}{-16x+10}\\ $ [What did we do here?] The process stops here because $x^2+5x+3$ is a polynomial of the second degree and $-16x+10$ is a polynomial of the first degree. So it follows that ${r(x)}={-16x+10}$, ${q(x)}={8x-5}$, and $ \dfrac{8x^3+35x^2-17x-5}{x^2+5x+3}={8x-5}+\dfrac{{-16x+10}}{x^2+5x+3}$ To conclude, $q(x)=8x-5$ $r(x)=-16x+10$